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\title{\heiti\zihao{2} 习题11.1}
\author{中书君}
\date{\songti \today}

\begin{document}
\maketitle
\section{求下列级数的和}
\subsection{$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n}}{2^{n}}$}
\textbf{解}\quad
记$S_{n}=\sum\limits_{k=1}^{n} \left(-\frac{1}{2}\right)^{n}$,则$S_{n}+\frac{1}{2}S_{n}=\frac{3}{2}S_{n}=-\frac{1}{2}-\left(-\frac{1}{2}\right)^{n+1}$.从而$S_{n}=\frac{2}{3}\cdot \mathrm{RHS}$

则有$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n}}{2^{n}}$=$\lim_{n \rightarrow \infty}S_{n}=-\frac{1}{3}$

\subsection{$\sum\limits_{n=1}^{\infty} \frac{1}{n(n+m)}$ (其中 $m$ 为一给定正整数)}
\textbf{解}\quad
$\sum\limits_{n=1}^{\infty} \frac{1}{n(n+m)}=\frac{1}{1+m}+\frac{1}{2+m}+\cdots=\frac{1}{m}\left[\left(\frac{1}{1}-\frac{1}{1+m}\right)+\left(\frac{1}{2}-\frac{1}{m+2}\right)+\cdots\right]=\frac{1}{m}\sum\limits_{i=1}^{m}\frac{1}{i}$

\subsection{$\sum\limits_{n=2}^{\infty} \frac{1}{n^{2}-1}$}
\textbf{解}\quad
$\sum\limits_{n=2}^{\infty} \frac{1}{n^{2}-1}=\sum\limits_{n=2}^{\infty}\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}\right)=\frac{3}{4}$

\subsection{$\sum\limits_{n=1}^{\infty} \frac{1}{(2 n-1)(2 n+1)}$}
\textbf{解}\quad
$\sum\limits_{n=1}^{\infty} \frac{1}{(2 n-1)(2 n+1)}=\sum\limits_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}$

\subsection{$\sum\limits_{n=1}^{\infty} \frac{1}{(3 n-2)(3 n+1)}$}
\textbf{解}\quad
$\sum\limits_{n=1}^{\infty} \frac{1}{(3 n-2)(3 n+1)}=\sum\limits_{n=1}^{\infty}\frac{1}{3}\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right)=\frac{1}{3}$

\subsection{$\sum\limits_{n=1}^{\infty} \frac{2 n-1}{2^{n}}$}
\textbf{解}\quad
记$S_{n}=\sum\limits_{k=1}^{n} \frac{2 k-1}{2^{k}}$,$\frac{1}{2}S_{n}=\sum\limits_{k=1}^{n} \frac{2 k-1}{2^{k+1}}$,从而$S_{n}=\frac{1}{2}+\sum\limits_{k=2}^{n} \frac{2}{2^{k}}-\frac{2n+1}{2^{n+1}}$,从而$\sum\limits_{n=1}^{\infty} \frac{2 n-1}{2^{n}}=\lim_{n \rightarrow \infty} S_{n}=\frac{1}{2}+\frac{5}{2}=3$

\subsection{$\sum\limits_{n=1}^{\infty} \frac{2 n+1}{n^{2}(n+1)^{2}}$}
\textbf{解}\quad
$\sum\limits_{n=1}^{\infty} \frac{2 n+1}{n^{2}(n+1)^{2}}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)=1$

\subsection{$\sum\limits_{n=1}^{\infty} \ln \frac{n(2 n+1)}{(n+1)(2 n-1)}$}
\textbf{解}\quad
设$S_{n}=\sum\limits_{k=1}^{n}\ln\frac{k(2 k+1)}{(k+1)(2 k-1)}=\ln (k) -\ln (k+1) + \ln (2k+1) - \ln (2k-1)=\ln(2n+1)-\ln(n+1)=\ln (\frac{2n+1}{n+1})$,从而$\sum\limits_{n=1}^{\infty} \ln \frac{n(2 n+1)}{(n+1)(2 n-1)}=\lim_{n \rightarrow \infty}S_{n}=\ln 2$

\subsection{$\sum\limits_{n=1}^{\infty}(\sqrt{n+2}-2 \sqrt{n+1}+\sqrt{n})$}
\textbf{解}\quad
$\sum\limits_{n=1}^{\infty}(\sqrt{n+2}-2 \sqrt{n+1}+\sqrt{n})=\sum\limits_{n=1}^{\infty}(\sqrt{n+2}-\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n})=\lim_{n \rightarrow \infty}\sqrt{n+2}-\sqrt{n+1}+1-\sqrt{2}=\lim_{n \rightarrow \infty}\frac{1}{\sqrt{n+2}+\sqrt{n+1}}+1-\sqrt{2}=1-\sqrt{2}$

\section{证明下列级数发散}
\subsection{$\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt[n]{n}}$}
\textbf{证}\quad
$\lim_{n\rightarrow \infty}\frac{1}{\sqrt[n]{n}}=1\neq 0$,故发散.

\subsection{$\sum\limits_{n=1}^{\infty}(-1)^{n}\frac{n}{n+1}$}
\textbf{证}\quad
$\lim_{n\rightarrow \infty}x_{n}=\lim_{n\rightarrow \infty}(-1)^{n}\frac{n}{n+1}\neq 0$,发散.

\subsection{$\sum\limits_{n=1}^{\infty}\left(1-\frac{1}{n}\right)^{n}$}
\textbf{证}\quad
$\lim_{n\rightarrow \infty}\left(1-\frac{1}{n}\right)^{n}=\frac{1}{e}\neq 0$,发散.

\subsection{$\sum\limits_{n=1}^{\infty}\frac{n-\ln n}{n+\sqrt{n}}$}
\textbf{证}\quad
$\lim_{n\rightarrow \infty}\frac{n-\ln n}{n+\sqrt{n}}=\lim_{n\rightarrow \infty}\frac{1-\frac{\ln n}{n}}{1+\frac{\sqrt{n}}{n}}=1\neq 0$,发散.

\section{设级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 收敛, $\sum\limits_{n=1}^{\infty} y_{n}$ 发散,证明:级数 $\sum\limits_{n=1}^{\infty}\left(x_{n}+y_{n}\right)$ 发散.}
\textbf{反证}\quad
若级数$\sum\limits_{n=1}^{\infty}\left(x_{n}+y_{n}\right)$ 收敛,则记其为$S$,记$\sum\limits_{n=1}^{\infty} x_{n}=A$,则$\sum\limits_{n=1}^{\infty} y_{n}$收敛到$S-A$,与$\sum\limits_{n=1}^{\infty} y_{n}$发散矛盾.证毕.

\section{设级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 收敛,证明:级数 $\sum\limits_{n=1}^{\infty}\left(x_{n}+x_{n+1}\right)$ 也收敛,并举例说明逆命题不成立.}
\textbf{证}\quad
$\sum\limits_{n=1}^{\infty}\left(x_{n}+x_{n+1}\right)=2\left(\sum\limits_{n=1}^{\infty} x_{n}\right)-x_{1}$,改变有限项,收敛.

对于$x_{n}=(-1)^{n}$有$\sum\limits_{n=1}^{\infty}\left(x_{n}+x_{n+1}\right)$收敛,而$\sum\limits_{n=1}^{\infty} x_{n}$不收敛.

\section{设数列 $\left\{n x_{n}\right\}$ 与级数 $\sum\limits_{n=1}^{\infty} n\left(x_{n}-x_{n+1}\right)$ 都收敛,证明 :级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 收敛.}
\textbf{证}\quad
$\sum\limits_{n=1}^{\infty} n\left(x_{n}-x_{n+1}\right)=\sum\limits_{n=1}^{\infty}x_{n}-\lim_{n\rightarrow \infty}nx_{n}$,从而$\sum\limits_{n=1}^{\infty} x_{n}=\sum\limits_{n=1}^{\infty} n\left(x_{n}-x_{n+1}\right)-\lim_{n\rightarrow \infty}nx_{n}$,收敛.

\section{已知级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 收敛,证明:
  $$
	  \lim _{n \rightarrow \infty} \frac{x_{1}+2 x_{2}+\cdots+n x_{n}}{n}=0
  $$}
\textbf{证}\quad
$\mathrm{Abel}$变换:令$a_{n}=n,b_{n}=\sum\limits_{k=1}^{n}x_{k}$,则
$$
	\begin{aligned}
        \frac{x_{1}+2 x_{2}+\cdots+n x_{n}}{n} & =\frac{\sum\limits_{k=1}^{n-1}(a_{k}-a_{k+1})b_{k}+a_{n}b_{n}}{n}\\
        &=b_{n}-\frac{\sum\limits_{k=1}^{n-1}b_{k}}{n}
	\end{aligned}
$$
取极限可得
$$
\lim_{n\rightarrow \infty}b_{n}-\frac{\sum\limits_{k=1}^{n-1}b_{k}}{n}=0
$$

关于$\mathrm{Abel}$变换可参考工科数学分析下册中的理论知识$\mathrm{Abel}$变换.或者回复Abel变换.


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